Lebesgue Measure: No Atoms! Disclaimer: This is just meant as record of a proof. For more details see: Answer own Question

How to prove that the Lebesgue measure has no atoms:
$$\lambda:\mathbb{R}^n\to\mathbb{R}_+$$
(Recall the definition for atoms: Measure Atoms: Definition?)
 A: Assuming, the Lebesgue measure has an atom $A\in\mathcal{B}(\mathbb{R}^n)$.
First consider growing boxes:
$$Q_k:=[-k,k]^n$$
By definition one of the following always applies:
$$\lambda(A\cap Q_k)=0\lor\lambda(A\cap Q_k^c)=0$$
If always the former case applies then there's a contradiction as $0=\mu(A\cap Q_k)\to\lambda(A)>0$.
So the atom splits into one within a finite box $A':=A\cap Q_K$.
Next consider cutting the box in halves:
$$Q_{K;i,\pm}:=[-K,K]^n\cap\left([-K,K]^{i-1}\times\mathbb{R}_\pm\times[-K,K]^{n-i}\right)$$
Then again one of the following applies:
$$\lambda(A'\cap Q_{K;i,+})=0\lor\lambda(A'\cap Q_{K;i,-})=0$$
Proceeding this way one obtains at a certain point a box of size:
$$\lambda(A)>\left(\frac{K}{2}\right)^L=\lambda(Q_{K;L})\leq\lambda(A'\cap Q_{K;L})=\lambda(A')=\lambda(A)\quad(\lambda(A)>0)$$
That is a contradiction!
Concluding, the Lebesgue measure has no atoms.
A: Here is one way: Let $B_r$ be the box $\{x\in\mathbb{R}^n: |x_k|\le r\text{ for } k=1,\ldots,n\}$. You can easily prove that $\lambda(B_r\cap A)$ is a continuous function of $r$, for any Lebesgue measurable set $A$. From this, it follows immediately that $A$ is not an atom.
Edit: To explain the last step, $\lambda(B_r\cap A)$ is a continuous function of $r$ taking the value $0$ at $r=0$ and $\lambda(A)$ in the limit $r\to\infty$. But if $A$ is an atom, this function can only take two values.
